Problem: Is ${179593}$ divisible by $9$ ?
Explanation: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {179593}= &&{1}\cdot100000+ \\&&{7}\cdot10000+ \\&&{9}\cdot1000+ \\&&{5}\cdot100+ \\&&{9}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {179593}= &&{1}(99999+1)+ \\&&{7}(9999+1)+ \\&&{9}(999+1)+ \\&&{5}(99+1)+ \\&&{9}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {179593}= &&\gray{1\cdot99999}+ \\&&\gray{7\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {1}+{7}+{9}+{5}+{9}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${179593}$ is divisible by $9$ if ${ 1}+{7}+{9}+{5}+{9}+{3}$ is divisible by $9$ Add the digits of ${179593}$ $ {1}+{7}+{9}+{5}+{9}+{3} = {34} $ If ${34}$ is divisible by $9$ , then ${179593}$ must also be divisible by $9$ ${34}$ is not divisible by $9$, therefore ${179593}$ must not be divisible by $9$.